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MathGroup Archive 2006

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Re: Problem with LaplaceTransform and InverseLaplaceTransform

  • To: mathgroup at smc.vnet.net
  • Subject: [mg67541] Re: Problem with LaplaceTransform and InverseLaplaceTransform
  • From: ab_def at prontomail.com
  • Date: Fri, 30 Jun 2006 04:14:26 -0400 (EDT)
  • References: <e7td5h$3kh$1@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

aXi wrote:
> Can someone point me to site that would help me in using Mathematica
> 5.2for purpose of calculations related to Systhems theory. I'm having
> trouble with several topics... for example doing
> InverseLaplaceTransform of function:
>
> (1/s) * (TanH[pi*s/2])
>
> Thanks in advance!

If you want the inverse Laplace transform of Tanh[Pi*s/2]/s, it can be
found as follows. First take the transform of Tanh[Pi*s/2]:

In[1]:= fs = Tanh[Pi*s/2]/s;

In[2]:= InverseLaplaceTransform[s*fs, s, t]

Out[2]= DiracDelta[t] + 2*Sum[(-1)^K$40*DiracDelta[(-K$40)*Pi + t],
{K$40, 1, Infinity}]

Then integrate the result:

In[3]:= Integrate[DiracDelta[t], t] +
  2*Sum[Integrate[(-1)^k*DiracDelta[t - Pi*k], t], {k, Infinity}]

Out[3]= UnitStep[t] + (-1 + (-1)^Floor[t/Pi])*UnitStep[-1 + t/Pi]

Strictly speaking, we need to evaluate Integrate[f[tau], {tau, 0-, t}],
'including' the zero in the integration range.

It is easy to see that the result is equal to (-1)^Floor[t/Pi] or
Sign[Sin[t]]. Here's a check:

In[4]:= LaplaceTransform[Sign[Sin[t]], t, s] - fs // Simplify

Out[4]= 0

Maxim Rytin
m.r at inbox.ru


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